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Cube-root function with horizontal asymptote y = 1 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
The real parameters mm and nn are such that the graph of the function f(x)=8x3+mx23−nxf\left(x\right) = \sqrt[3]{8x^3+mx^2}-nx has the horizontal asymptote y=1y = 1, then
A∣m+n∣=10\left|m+n\right| = 10
B∣m+n∣=14\left|m+n\right| = 14correct
C∣m−n∣=10\left|m-n\right| = 10correct
D∣m−n∣=2\left|m-n\right| = 2
Solution
Step 1: A horizontal asymptote y=1y = 1 means
lim⁡x→∞(8x3+mx23−nx)=1\lim_{x\to\infty}\left(\sqrt[3]{8x^3+mx^2}-nx\right) = 1
 Step 2: With p=8x3+mx23p = \sqrt[3]{8x^3+mx^2}, q=nxq = nx and p−q=p3−q3p2+pq+q2p-q = \dfrac{p^3-q^3}{p^2+pq+q^2},
f(x)=(8x3+mx2)−n3x3(8x3+mx23)2+nx8x3+mx23+n2x2=(8−n3)x3+mx2(8x3+mx23)2+nx8x3+mx23+n2x2f(x) = \frac{\left(8x^3+mx^2\right)-n^3x^3} {\left(\sqrt[3]{8x^3+mx^2}\right)^2+nx\sqrt[3]{8x^3+mx^2}+n^2x^2} = \frac{\left(8-n^3\right)x^3+mx^2} {\left(\sqrt[3]{8x^3+mx^2}\right)^2+nx\sqrt[3]{8x^3+mx^2}+n^2x^2}
 Step 3: The denominator grows like x2x^2. If 8−n3≠08-n^3 \ne 0 the numerator grows like x3x^3 and the quotient →±∞\to \pm\infty, which is not possible. So
8−n3=0⟹n=28-n^3 = 0 \quad\Longrightarrow\quad n = 2
 nn is real and n3=8n^3 = 8 has only one real root, so n=2n = 2 is the only value. Step 4: With n=2n = 2, dividing numerator and denominator by x2x^2,
f(x)=m(8+mx)23+28+mx3+4 →x→∞ m643+283+4=m4+4+4=m12f(x) = \frac{m}{\sqrt[3]{\left(8+\dfrac mx\right)^2}+2\sqrt[3]{8+\dfrac mx}+4} \ \xrightarrow[x\to\infty]{}\ \frac{m}{\sqrt[3]{64}+2\sqrt[3]8+4} = \frac{m}{4+4+4} = \frac{m}{12}
m12=1⟹m=12\frac{m}{12} = 1 \quad\Longrightarrow\quad m = 12
 So (m,n)=(12,2)\left(m,n\right) = \left(12,2\right). The same pair works as x→−∞x\to-\infty, since the real cube root keeps the sign of its argument, so 8x3+⋯3∼2x\sqrt[3]{8x^3+\cdots}\sim 2x there too. Step 5:
∣m+n∣=∣14∣=14,∣m−n∣=∣10∣=10\left|m+n\right| = \left|14\right| = 14 ,\qquad \left|m-n\right| = \left|10\right| = 10
 So (2) and (3) hold, while (A) claims ∣m+n∣=10\left|m+n\right| = 10 and (4) claims ∣m−n∣=2\left|m-n\right| = 2, both false. Answer: (2) and (3)
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